# Thread: Differentiation using first principles

1. ## Differentiation using first principles

I'm currently studying a chapter on Differentiation. I was introduced to first prinicples, and i found a few of the first questions easy. But, ive hit a dead end, and nothing I try with this next question works. I looked up the method and found explanations at:
First Principles

The question goes:

"Use First principles to find $\displaystyle dy/dx$

$\displaystyle y=6x-2x^2$

Now i used the formula, and got something like this:

$\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\displaystyle = \lim_{h \rightarrow 0} \frac{6(x+h)-2(x+h)^2-(6x-2x^2)}{h}$

From here, I have no idea, Ive tried many different ways and I always end up with the wrong answer.

Please explain to me how to do this.

2. Ok, so we have $\displaystyle y = -2x^{2} + 6x$, so
$\displaystyle f(x+h) = -2(x+h)^{2} + 6(x+h)$
we'll expand this, such that we get:
$\displaystyle -2x{^2} -4xh - 2h{^2} + 6x + 6h$

We now need to find:
$\displaystyle dy/dx = \lim_{h->0} [f(x+h) - f(x)] \div (h)$

$\displaystyle = \lim_{h->0} [(-2x^{2} - 4xh - 2x^{2} + 6x + 6h) - (-2x^{2} + 6x)] \div (h)$

$\displaystyle = \lim_{h->0} h(-2h - 4x + 6)\div h$

$\displaystyle = \lim_{h->0} (-2h -4x + 6)$

$\displaystyle = -4x + 6$

3. Originally Posted by enjam
$\displaystyle = \lim_{h->0} [(-2x^{2} - 4xh - 2x^{2} + 6x + 6h) - (-2x^{2} + 6x)] \div (h)$

$\displaystyle = \lim_{h->0} h(-2h - 4x + 6)\div h$
Could you please tell me how you get from the first part of this to the second?

4. Sorry, about that made a small typo, it's actually meant to be this:
$\displaystyle \lim_{h->0} [(-2x^{2} - 4xh - 2h^{2} + 6x + 6h) - (-2x^{2} + 6x)] \div (h)$

So we have -2x^2 + 2x^2 + 6x - 6x - 4xh - 2h^2 + 6h

= -4xh -2h^2 + 6h
= h(-4x - 2h + 6)